For the next step we need to do some math. We need to be able to determine which one of the 4 "push" segments a point is actually in. There's a unit test to check our work, but now we need to work out the math to make it happen.

Let's review some basic principles. The Squeak graphics coordinate system has its origin in the upper left corner of the display. As you travel across the screen to the right your X values increase. As you travel across the screen heading down your Y values increase. This can be a little confusing since it's inverted for the Y-axis from what you normally think of when drawing lines.

I'm going to flip over our little cross-hatch diagram to make it easier to keep the math straight.

The push regions end up like this for our flipped over diagram. We haven't really changed anything here. The North push side is still on the side farthest from the 0 position in the y-axis.

There are 2 line segments in our diagram. One that begins in the lower left corner and heads up and to the right and the other that begins in the upper left corner and heads down and to the right. Let's call the first line our "heading-up" line and the second line our "heading down" line.

A student of this tutorial wrote to me
and mentioned that the diagram
with the 4 compass points looked incorrect, and at the least was confusing.
Actually, the diagram is correct.

Here's what's going on. The reason this looks incorrect has to do with the behavior
that is being calculated. At this point in the tutorial the algorithm needs to calculate
which direction the mirror would be pushed if user clicks the mouse inside the cube.

For example, if the user clicks on the left side of the cube the mirror should be
pushed East. If clicked on the right side, the mirror should be pushed West.

The really confusing part is that, for the diagrams, if the user clicks in the
lower part of the cube we want the mirror to move North. However, as stated in
the text before the diagrams, Smalltalk has an inverted Y-axis. So for the math
to work right, the click for pushing North is drawn in the "upper" portion of the
cube and "South" in the lower portion.

Continuing, recall that the equation for a straight line is

y = mx + b

Let's solve for the values of the "heading=up" line equation.

x1 = 10

y1 = 10

x2 = 20

y2 = 20

dy = y2 - y1

dy = 20 - 10

dy = 10

dx = x2 - x1

dx = 20 - 10

dx = 10

m = dy / dx

m = 10 / 10

m = 1

To solve for "b" we plug in one of our known points.

y = mx + b

x = 10

y = 10

10 = m(10) + b

m = 1

10 = 1(10) + b

10 - 10 = b

b = 0

The equation for our "heading-up" line is

y = mx + b

y = 1(x) + 0

y = x

Now we solve for the equation of our "heading-down" line.

x1 = 10

y1 = 20

x2 = 20

y2 = 10

dy = y2 - y1

dy = 10 - 20

dy = -10

dx = x2 - x1

dx = 20 - 10

dx = 10

m = dy / dx

m = -10 / 10

m = -1

To solve for "b" we once again plug in one of our known points.

y = mx + b

x = 10

y = 20

20 = m(10) + b

m = -1

20 = -1(10) + b

20 + 10 = b

b = 30

The equation for our "heading-down" line is

y = mx + b

y = -1(x) + 30

y = 30 - x